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8. Sequences and Series
hard
If $1, \log _{10}\left(4^{x}-2\right)$ and $\log _{10}\left(4^{x}+\frac{18}{5}\right)$ are in
arithmetic progression for a real number $x$ then the value of the determinant $\left|\begin{array}{ccc}2\left(x-\frac{1}{2}\right) & x-1 & x^{2} \\ 1 & 0 & x \\ x & 1 & 0\end{array}\right|$ is equal to ...... .
A
$5$
B
$4$
C
$1$
D
$2$
(JEE MAIN-2021)
Solution
$2 \log _{10}\left(4^{ x }-2\right)=1+\log _{10}\left(4^{ x }+\frac{18}{5}\right)$
$\left(4^{ x }-2\right)^{2}=10\left(4^{ x }+\frac{18}{5}\right)$
$\left(4^{ x }\right)^{2}+4-4\left(4^{ x }\right)-32=0$
$\left(4^{ x }-16\right)\left(4^{ x }+2\right)=0$
$4^{ x }=16$
$x =2$
$\left|\begin{array}{lll}3 & 1 & 4 \\ 1 & 0 & 2 \\ 2 & 1 & 0\end{array}\right|=3(-2)-1(0-4)+4(1)$
$=-6+4+4=2$
Standard 11
Mathematics
